A triangle has a base of 16 inches and an altitude of 8 inches. Find the dimentions of the largest rectangle that can be inscribed in the triangle if the base of the rectangle coincides with the base of the triangle
Check your post's wording.
Is the triangle isosceles?
To find the dimensions of the largest rectangle that can be inscribed in the given triangle, we can start by visualizing the problem.
We know that the base of the triangle is 16 inches, and the altitude is 8 inches. Let's call the dimensions of the rectangle length (L) and width (W).
Since the base of the rectangle coincides with the base of the triangle, the length of the rectangle is equal to the base of the triangle, which is 16 inches.
To maximize the area of the rectangle, we need to find the maximum width that can fit within the triangle.
To do this, we consider that any line drawn from one vertex to the opposite side in a triangle is perpendicular to that side. In this case, we can draw a line from the top vertex of the triangle to the base, creating a right triangle.
The altitude of the triangle is the height of the right triangle we created. Let's label that as H.
Now, we have a right triangle with a base of 16 inches, a height of 8 inches, and a hypotenuse of H.
Using the Pythagorean theorem, we can find H:
H^2 = (base)^2 + (height)^2
H^2 = 16^2 + 8^2
H^2 = 256 + 64
H^2 = 320
H = sqrt(320)
H = 4sqrt(5) inches
Since the width of the rectangle cannot exceed the height (H) of the right triangle, the width is limited to a maximum of 4sqrt(5) inches.
Therefore, the dimensions of the largest rectangle that can be inscribed in the triangle are:
Length (L) = 16 inches
Width (W) = 4sqrt(5) inches